# Factoring (Again)

1. Mar 2, 2008

### Andy111

1. The problem statement, all variables and given/known data
I had a topic somewhere about factoring, and now I have some more factoring problems I don't understand.

such as:

1)(x-1)$$^{3}$$ - (x+2)$$^{3}$$

2)64x$$^{3}$$ - 27y$$^{3}$$

3)3ab - 20cd -15ac + 4bd

2. Relevant equations

3. The attempt at a solution

2. Mar 2, 2008

### rocomath

1. If I use substitutions, it may make things a little easier.

$$a=(x-1)$$
$$b=(x+2)$$

$$a^3-b^3=(a-b)(a^2+ab+b^2)$$

$$(x-1)^3-(x+2)^3=[(x-1)-(x+2)][(x-1)^2+(x-1)(x+2)+(x+2)^2]$$
$$=(x-1-x-2)(x^2-2x+1+x^2+x-2+x^2+4x+4)$$

Does this confuse you more? Continue simplifying and collecting like terms and it's solved.

2. Can you re-write it?

What number must you raise to the power of 3, to attain 64 and 27? You want to choose a number so you can raise both your coefficient and variable to the same power.

3. Factor by grouping.

Last edited: Mar 2, 2008
3. Mar 3, 2008

### Andy111

No, that's not confusing, I understand what you're doing in 1.

I know 4 cubed and 3 cubed = 64 and 27, but I don't know how it looks in factored form.

4. Mar 3, 2008

### HallsofIvy

Staff Emeritus
Then you should understand that rocophyics just told you what it looks like in factored form: x3- y3= (x- y)(x2+ xy+ y2).

Last edited: Mar 3, 2008
5. Mar 3, 2008

### tiny-tim

Hi Andy!

64x$$^{3}$$ - 27y$$^{3}$$ = 4$$^{3}$$x$$^{3}$$ - 3$$^{3}$$y$$^{3}$$: does that help?

6. Mar 3, 2008

### rocomath

$$a^mb^m=(ab)^m$$

7. Mar 3, 2008

### Andy111

Oh, okay, so 2 is like 1, but with coefficients.

8. Mar 3, 2008

### tiny-tim

You've got 1) and 2) now.

Have you got 3)?

If not, there are various ways of doing it - one is to write it as a 2x2 matrix.

[size=-2](if you're ok now, don't forget to mark thread "solved"!)[/size]​

9. Mar 3, 2008

### Andy111

Yeah, I got 3,

I believe this is right (3a + 4d)(b - 5c).

I understand the formula for x^3 - y^3, but what the minus sign is instead a plus sign?

Last edited: Mar 3, 2008
10. Mar 3, 2008

### rocomath

$$(x+y)(x^2-xy+y^2)$$

Replace y, with (-y) and tell me what your new equation is.

11. Mar 4, 2008

### tiny-tim

factoring x^3 + y^3 ?

Yay!

Hint: divide by y^3, so you get (x/y)^3 + 1.

Put x/y = z, so you get z^3 + 1.

Can you see how to factor that?

(if not, come back for another hint)

12. Mar 4, 2008

### HallsofIvy

Staff Emeritus
$$x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ x^{n-3}y^2\cdot\cdot\cdot+ x^2y^{n-3}+ xy^{n-2}+ y^{n-1}$$
for n any positive integer.

$$x^n+ y^n= (x+ y)(x^{n-1}- x^{n-2}y+ x^{n-2}y^2\cdot\cdot\cdot- x^2y^{n-3}+ xy^{n-2}- y^{n-1}$$
for n any odd integer.